A nucleus with mass number 220 is initially a | vedclass

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(B) Let the mass number of the parent nucleus be $A = 220$. After emitting an $\alpha$-particle $(m_2 = 4)$,the daughter nucleus has a mass number $m_

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$5.4$

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(B) Let the mass number of the parent nucleus be $A = 220$. After emitting an $\alpha$-particle $(m_2 = 4)$,the daughter nucleus has a mass number $m_1 = 220 - 4 = 216$.The total energy released in the reaction is $Q = K_1 + K_2 = 5.5 \, MeV$,where $K_1$ is the kinetic energy of the daughter nucleus and $K_2$ is the kinetic energy of the $\alpha$-particle.By the law of conservation of linear momentum,the initial momentum is zero,so the magnitudes of the momenta of the daughter nucleus and the $\alpha$-particle must be equal:$p_1 = p_2$$\sqrt{2 m_1 K_1} = \sqrt{2 m_2 K_2}$$m_1 K_1 = m_2 K_2$$216 K_1 = 4 K_2$$K_1 = \frac{4}{216} K_2 = \frac{1}{54} K_2$Substituting this into the $Q$-value equation:$K_1 + K_2 = 5.5$$\frac{1}{54} K_2 + K_2 = 5.5$$\frac{55}{54} K_2 = 5.5$$K_2 = 5.5 	imes \frac{54}{55} = 0.1 	imes 54 = 5.4 \, MeV$.Thus,the kinetic energy of the $\alpha$-particle is $5.4 \, MeV$.

$A$ nucleus with mass number $220$ is initially at rest. It emits an $\alpha$-particle. If the $Q$-value of the reaction is $5.5 \, MeV$,then the kinetic energy of the $\alpha$-particle will be .......... $MeV$.

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